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MA280 Probability
1
Overview
Probability theory:
a tool of fundamental importance to nearly all scientists,
engineers, medical practitioners jurists, and industrialists
This course provides an elementary introduction to the
theory of probability
2
The basic principle of counting
Situation:
Two experiments to be performed
Experiment 1: can result in any of m possible outcomes
For each outcome of experiment 1, experiment 2 can result
in any of the n possible outcomes.
Conclusion:
The total number of possible outcomes of the two
experiments together: mn
Proof. List all the possible outcomes of the 2 experiments.
3
Example 2a
A small community consists of 10 women, each has 3 children.
If one woman and one of her children are to be chosen as
mother and child of the year, how many different choices are
possible?
4
The generalized basic principle of counting
Situation
r experiments are to be performed.
Experiment 1 may result in n1 possible outcomes
For each of the possible outcomes for Experiment 1,
Experiment 2 may result in n2 possible outcomes
For each of the possible outcomes for Experiments 1 and
2, Experiment 3 may result in n3 possible outcomes
. . .
Conclusion
The total number of possible outcomes for the r
experiments together: n1 · n2 · · · nr .
5
Example 2b
A college planning committee consists of 3 freshmen, 4
sophomores, 5 juniors, and 2 seniors. A subcommittee of 4,
consisting of 1 person from each class, is to be chosen. How
many different subcommittees are possible?
6
Notation: !
Suppose there are n objects.
The total number of different permutations (ordered
arrangements) of the n objects:
n! = n(n − 1)(n − 2) · · · 3 · 2 · 1
Notes:
0! := 1
n! reads as “n factorial”
7
Example 3a
How many different batting orders are possible for a baseball
team consisting of 9 players?
Solution. 9! = 9 ∗ 8 ∗ · · · ∗ 1 = 362,880.
8
Permutations when certain of the objects are
indistinguishable from one another
Example 3d How many different letter arrangements can be
formed from the letters pepper?
Solution.6!/(3!2!)
There are 6! permutations of the letters when the 3p’s and the
2e’s are distinguished from one another.
However, consider any one of these permutations for instance,
if we now permute the p’s among themselves and the e’s
among themselves,
then the resultant 3!2! arrangements would still be of the same
form when the same letters are indistinguishable from one
another
Activity: Explain the above idea to each other.
9
Permutations when certain of the objects are
indistinguishable from one another
The total number of different permutations of n objects, of
which n1 are alike, n2 are alike, · · · , nr are alike is:
n!
n1!n2! · · · nr !
10
Exercise
A chess tournament has 10 competitors, of which 4 are
Russian, 3 are from the United States, 2 are from Great Britain,
and 1 is from Brazil. If the tournament result lists just the
nationalities of the players in the order in which they placed,
how many outcomes are possible?